A massless rigid rod $OP$ of length $L$ is hinged frictionlessly at $O$. A concentrated mass $m$ is attached to end $P$ of the rod. Initially, the rod $OP$ is horizontal. Then, it is released from rest. There is gravity as shown. The rod acquires an angular velocity as it swings. The clockwise angular velocity of the rod, when it first reaches the vertical position as shown, is:
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For pendulum-like systems with a point mass at the end of a massless rod, always use energy conservation. The drop in height directly gives the rotational kinetic energy at the bottom.
Step 1: Identify the physical principle.
The system consists of a rigid rod hinged at $O$ with a point mass $m$ at its end $P$. Since the rod is massless, only the point mass contributes to the dynamics. The motion is governed by conservation of mechanical energy:
\[
\text{Loss in potential energy} = \text{Gain in rotational kinetic energy}
\]
Step 2: Calculate change in potential energy.
- Initially, mass $m$ is at horizontal distance $L$ from $O$ at the same vertical level as $O$.
- When the rod swings down to the vertical, the mass descends by a vertical distance equal to $L$.
Thus, the loss in potential energy is:
\[
\Delta U = mgL
\]
Step 3: Express rotational kinetic energy.
Rotational kinetic energy is:
\[
K = \frac{1}{2} I \omega^2
\]
where $I$ is the moment of inertia of the system about hinge $O$.
Since the rod is massless and all mass $m$ is concentrated at distance $L$:
\[
I = mL^2
\]
Step 4: Apply energy conservation.
\[
mgL = \frac{1}{2} (mL^2) \omega^2
\]
Step 5: Simplify.
\[
mgL = \frac{1}{2} mL^2 \omega^2
\]
Cancel $m$:
\[
gL = \frac{1}{2} L^2 \omega^2
\]
\[
\omega^2 = \frac{2g}{L}
\]
\[
\omega = \sqrt{\frac{2g}{L}}
\]
Step 6: Nature of motion.
Since the rod is released from rest and falls under gravity, the angular velocity at the vertical position is clockwise.
Final Answer:
\[
\boxed{\sqrt{\dfrac{2g}{L}}}
\]
